VOC Dilution Factor

Discussions about sample preparation: extraction, cleanup, derivatization, etc.

18 posts Page 1 of 2
Hi!

I am working on developing a method for quant some volatile organic compounds on GCMS. My way is like that;

For example; benzen limit is 10 ppm. I prepare 2,5 - 5 - 10 - 25 - 50 ppm calibration point.

We taking organic compounds with a activated charcoal tube and its have 150 mg activated carbon in it.

I break the tube and take the activated charcoal on a tube and add 5 mL of methanol (my solvent that i am using for all method). Mix them for 30 min. After 30 min. i take 1 mL to gc vial from 5 mL solution and give the GCMS to read.

Then when i quant the sample is my dilution factor 5/150 or 5/1 ?
Learn your math! The answer is;

(Conc. of sample in ug/mL) x (5 mL/1 mL) / (0.150 g / 1000 g per Kg)

Thus, the answer is in ug/Kg
An alternative is to report it ug/tube assuming you extract the all the charcoal in the tube. Then you need to know how much sample the tube represents to get to ppm.
As Steve said, if you are looking for concentration in a air sample pulled through the charcoal tube, then you find the mg/tube and the volume of air which may be 50 cubic meters. If you had 10mg per tube then you would have 10mg/50cubic meters or 0.2mg/cubic meter.

If it was a water sample passed though the charcoal then you use the volume of water.
The past is there to guide us into the future, not to dwell in.
ucanmoruk wrote:
...
I break the tube and take the activated charcoal on a tube and add 5 mL of methanol (my solvent that i am using for all method). Mix them for 30 min. After 30 min. i take 1 mL to gc vial from 5 mL solution and give the GCMS to read.

Then when i quant the sample is my dilution factor 5/150 or 5/1 ?

You absolutely must determine desorption rates for all your analytes and include them in calculations if they substantially differ from 100%.

In fact adsorption rate on charcoal from air stream should be investigated as well.

PS - your posts better fit to Student Projects section of this forum
dblux_ wrote:
ucanmoruk wrote:
PS - your posts better fit to Student Projects section of this forum


I'm so sory about the simple and absurd question for experts like you and the other forum members.

But i have some troubles on preparing VOC calibration before the working sample on tube. I don't have VOC standart mix and therefore i am preparing standarts by myself.

Look. I am looking for benzene, ethanol and isoamylalcohol. The limits in order for that compounds 0.038, 22.8 and 4.32 mg/m3 for 0.012 m3 activated charcoal tube.

My solvent is methanol. If i take 1 ml ethanol in 10 ml methanol its equal to 78,81 mg/m3 (from d=m/v)
take 1 ul in 10 ml meth = 0,088 mg/m3 benzene
take 200 ul in 10 ml meth = 16,184 mg/m3 isoamyl alcohol

The 10 ml mix solution my 5th point of the calibration. Then i dilute the mix solution 1:2:2:2.5:2 rate. Thats my calibration points.

After that; i take the tube sample, break it into 5 ml methanol. Shake it for about 30 minutes. Take 1 ml to gc vial and give it to gcms.

What is my wrong or missing point?
ucanmoruk wrote:
Look. I am looking for benzene, ethanol and isoamylalcohol. The limits in order for that compounds 0.038, 22.8 and 4.32 mg/m3 for 0.012 m3 activated charcoal tube.

You need to clarify. So far I guess you are going to determine benzene, ethanol and isoamyl alcohol concentration in air. What are these limits ? Are they max concentrations expected in air ? Are you sure that the volume of charcoal in the tube is 0,012 m3 (it's 12 L)? Typical charcoal tube has 2 layers of charcoal with total mass 150 mg.
My solvent is methanol. If i take 1 ml ethanol in 10 ml methanol its equal to 78,81 mg/m3 (from d=m/v)

If you dilute 1 mL of ethanol in methanol and fill up with diluent to 10 mL then ethanol concentration would be 78900 g/m3.
Before we go any further we have to clarify this.
dblux_ wrote:
You need to clarify. So far I guess you are going to determine benzene, ethanol and isoamyl alcohol concentration in air. What are these limits ? Are they max concentrations expected in air ? Are you sure that the volume of charcoal in the tube is 0,012 m3 (it's 12 L)? Typical charcoal tube has 2 layers of charcoal with total mass 150 mg.


Benzene limit is 0.038 mg/m3
İsoamyl alcohol limit is 4.32 mg/m3
Ethanol limit is 22.8 mg/m3

The limits are for 0.012 m3 air. Because we are collecting to charcoal tube about 0,012 m3 (it's 12 L) air. And yes charcoal tube has 2 layers of charcoal with total mass 150 mg.

dblux_ wrote:
If you dilute 1 mL of ethanol in methanol and fill up with diluent to 10 mL then ethanol concentration would be 78900 g/m3.
Before we go any further we have to clarify this.


Yes that's true. Ethanol concentration is 78900 g/m3 so 78.9 mg/m3. My reporting unit is mg/m3
ucanmoruk wrote:
Yes that's true. Ethanol concentration is 78900 g/m3 so 78.9 mg/m3. My reporting unit is mg/m3


78900 g/m3 equals to 78900000 mg/m3. Sorry, these are really the basics.
Peter Apps
dblux_ wrote:
ucanmoruk wrote:
Yes that's true. Ethanol concentration is 78900 g/m3 so 78.9 mg/m3. My reporting unit is mg/m3


78900 g/m3 equals to 78900000 mg/m3. Sorry, these are really the basics.


Sorry, thats my fault. I am not fine :)

Ethanol d: 789 kg/m3
Solvent v: 10 ml
I take from ethanol 1 ml = 1000 ul
1 m3 = 1000 L

789 kg/m3 = 789 mg/ml

(1 ml ethanol * 789 mg/ml) / 10 ml solvent = 78.9 mg

If i take 1 ml ethanol to 10 ml solvent, i have 78.9 mg ethanol on this solution
ucanmoruk wrote:
...
Ethanol d: 789 kg/m3
Solvent v: 10 ml
I take from ethanol 1 ml = 1000 ul
1 m3 = 1000 L

789 kg/m3 = 789 mg/ml

(1 ml ethanol * 789 mg/ml) / 10 ml solvent = 78.9 mg

If i take 1 ml ethanol to 10 ml solvent, i have 78.9 mg ethanol on this solution

Not exactly. If you add 1 mL of EtOH to 10 mL of MetOH you will get ca. 11 mL of solution not 10 mL. Probably you are going to add 1 mL of ethanol to MetOH and dilute it to 10 mL in volumetric flask.

Nevertheless, the concentration of solution will be 78.9 mg/mL.

PS - when you handle small amounts of substances it's better to use mL and mg instead of m3 and kg.
ucanmoruk wrote:
...
Ethanol d: 789 kg/m3
Solvent v: 10 ml
I take from ethanol 1 ml = 1000 ul
1 m3 = 1000 L

789 kg/m3 = 789 mg/ml

(1 ml ethanol * 789 mg/ml) / 10 ml solvent = 78.9 mg

If i take 1 ml ethanol to 10 ml solvent, i have 78.9 mg ethanol on this solution

[/quote]

No - you have 789 mg of ethanol in this solution. You put in 1 ml of ethanol, 1 ml of ethanol has a mass of 789 mg.

You seem to be confused about mass and concentration; a mass of 789 mg in 10 ml of solution gives a concentration of 78.9 mg/ml.

Peter
Peter Apps
Peter, it would be very kind if you modify quotation in your post in such a way that you ommit the words "dblux_ wrote:".
I simply don't want to be connectet in any way with such ignorance with concentration calculations.
dblux_ wrote:
Peter, it would be very kind if you modify quotation in your post in such a way that you ommit the words "dblux_ wrote:".
I simply don't want to be connectet in any way with such ignorance with concentration calculations.


Done
Peter Apps
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